Worked solution · Exam P

SOA Exam P Sample Question #680 Solution

This Exam P sample reference tests Exponential Distribution. This is a shifted-exponential percentile calculation. Setting the CDF to 0.10 gives a standardized excess of log(10/9), so the percentile is d+beta log(10/9), which is choice D.

Independent analysis by ActuaryProof · reviewed July 30, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

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Independent solution

How to solve this Exponential Distribution question

Setup

Express the supported variable as its lower endpoint plus an exponential excess with scale beta.

X=d+Y,\qquad Y\sim\operatorname{Exp}(\text{scale }\beta)

F_X(x)=1-e^{-(x-d)/\beta},\qquad x\ge d

Model

Let q denote the 10th percentile and set the shifted-exponential CDF at q equal to 0.10.

1-e^{-(q-d)/\beta}=0.10

e^{-(q-d)/\beta}=0.90

Compute

Take logarithms and solve for the percentile on the original scale.

\frac{q-d}{\beta}=-\log(0.90)=\log\!\left(\frac{10}{9}\right)

q=d+\beta\log\!\left(\frac{10}{9}\right)

Answer

The fixed shift must remain in the percentile, and beta multiplies the dimensionless logarithm.

\boxed{q_{0.10}=d+\beta\log(10/9)\quad\text{(D)}}

Test the method on unseen problems

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30 original univariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

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Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThis expression omits the location shift d and also uses log(11/10), which does not come from the required survival probability 0.90.
BThe logarithmic excess beta log(10/9) is correct, but it is measured above the lower endpoint; omitting d leaves the answer on the wrong origin.
CAdding d fixes the origin, but log(11/10) treats the 10% probability through an incorrect odds-style ratio instead of using -log(0.90).
EPlacing beta in a logarithmic power or denominator does not invert (q-d)/beta and also fails to retain the location shift.

Original practice · fully worked

Original variant: sensor recalibration latency

A remote sensor's recalibration latency equals a fixed 3-second handshake plus an independent exponential processing time with mean 8 seconds. Calculate the 75th percentile of the total latency.

A
2.301
B
5.301
C
11.090
D
14.090
E
35.000

Answer and calculation

At the 75th percentile, the exponential processing time has survival probability 0.25. Adding the fixed handshake to 8 log(4) gives 14.090355 seconds, so choice D is correct.

Setup

Write total latency T as the fixed handshake plus exponential processing time Y.

T=3+Y,\qquad Y\sim\operatorname{Exp}(\text{scale }8)

Model

For the 75th percentile q, the remaining upper-tail probability is one-quarter.

\Pr(T>q)=e^{-(q-3)/8}=0.25

Compute

Take logarithms, scale by the exponential mean, and restore the fixed delay.

\frac{q-3}{8}=-\log(0.25)=\log 4

q=3+8\log 4=14.0903548890\ldots

Answer

The 75th percentile of total latency is approximately 14.090 seconds.

\boxed{q_{0.75}\approx 14.090\quad\text{(D)}}

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Next steps for Exam P Univariate Random Variables

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How to solve this Exponential Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: sensor recalibration latency

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables